<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:
![\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})](https://tex.z-dn.net/?f=%5Cdisplaystyle%20500%20%5C%20g%20%5C%20C_6H_%7B12%7DO_6%28%5Cfrac%7B1%20%5C%20mol%20%5C%20C_6H_%7B12%7DO_6%7D%7B180.18%20%5C%20g%20%5C%20C_6H_%7B12%7DO_6%7D%29%28%5Cfrac%7B2%20%5C%20mol%20%5C%20C_2H_5OH%7D%7B1%20%5C%20mol%20%5C%20C_6H_%7B12%7DO_6%7D%29)
- [DA} Multiply/Divide [Cancel out units]:
![\displaystyle 5.55001 \ mol \ C_2H_5OH](https://tex.z-dn.net/?f=%5Cdisplaystyle%205.55001%20%5C%20mol%20%5C%20C_2H_5OH)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
Picoliter is a unit of measurement for liquids.
One picoliter = 1×10⁻⁹ mililiters
So:
19 mL ---- x pL
1×10⁻⁹ mL ---- 1 pL
1×10⁻⁹x = 19
x = 1,9 × 10¹⁰ pL
or 19,000,000,000 pL
Answer: 1.9 × 10¹⁰ pL
As the gas cools it condenses and becomes a liquid its atoms also become smaller
Answer:
The absorbance of the myoglobin solution across a 1 cm path is 0.84.
Explanation:
Beer-Lambert's law :
Formula used :
![A=\epsilon \times c\times l](https://tex.z-dn.net/?f=A%3D%5Cepsilon%20%5Ctimes%20c%5Ctimes%20l)
![A=\log \frac{I_o}{I}](https://tex.z-dn.net/?f=A%3D%5Clog%20%5Cfrac%7BI_o%7D%7BI%7D)
![\log \frac{I_o}{I}=\epsilon \times c\times l](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BI_o%7D%7BI%7D%3D%5Cepsilon%20%5Ctimes%20c%5Ctimes%20l)
where,
A = absorbance of solution
c = concentration of solution
= Molar absorption coefficient
l = path length
= incident light
= transmitted light
Given :
l = 1 cm, c = 1 mg/mL ,![\epsilon = 15,000 M^{-1}cm^{-1}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%2015%2C000%20M%5E%7B-1%7Dcm%5E%7B-1%7D)
Molar mass of myoglobin = 17.8 kDa = 17.8 kg/mol=17800 g/mol
(1 Da = 1 g/mol)
c = 1 mg /mL = ![{1mg /mL}{\text{Molar mass of myoglobin}}](https://tex.z-dn.net/?f=%20%7B1mg%20%2FmL%7D%7B%5Ctext%7BMolar%20mass%20of%20myoglobin%7D%7D)
![c = \frac{1 mg/mL}{ 17800 g/mol} = 5.6179\times 10^{-5} mol/L](https://tex.z-dn.net/?f=c%20%3D%20%5Cfrac%7B1%20mg%2FmL%7D%7B%2017800%20g%2Fmol%7D%20%3D%205.6179%5Ctimes%2010%5E%7B-5%7D%20mol%2FL)
1 mg = 0.001 g, 1 mL = 0.001 L
![A= 15,000 M^{-1}cm^{-1}\times 5.6179\times 10^{-5} mol/L\times 1 cm](https://tex.z-dn.net/?f=A%3D%2015%2C000%20M%5E%7B-1%7Dcm%5E%7B-1%7D%5Ctimes%205.6179%5Ctimes%2010%5E%7B-5%7D%20mol%2FL%5Ctimes%201%20cm)
![A=0.8426 \approx 0.84](https://tex.z-dn.net/?f=A%3D0.8426%20%5Capprox%200.84)
The absorbance of the myoglobin solution across a 1 cm path is 0.84.